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Three equal masses of 1 kg each are plac...

Three equal masses of 1 kg each are placed at the vertices of an equilateral triangle PQR and a mass of 2 kg is placed at the centroid O of the triangle which is at a distance of `sqrt(2)m` from each of the vertices of the triangle. The force, in newton, acting on the mass of 2 kg is :

A

2

B

1

C

3

D

zero

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The correct Answer is:
To solve the problem, we need to calculate the gravitational force acting on the 2 kg mass placed at the centroid O of the equilateral triangle formed by the three 1 kg masses at vertices P, Q, and R. ### Step-by-Step Solution: 1. **Identify the Forces Acting on the 2 kg Mass**: The three 1 kg masses at the vertices P, Q, and R exert gravitational forces on the 2 kg mass at the centroid O. The gravitational force between two masses is given by the formula: \[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \] where \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses, and \( r \) is the distance between them. 2. **Calculate the Force from One Mass**: For each of the 1 kg masses (let's denote one of them as \( m_1 = 1 \, \text{kg} \)) acting on the 2 kg mass (\( m_2 = 2 \, \text{kg} \)), and given that the distance \( r = \sqrt{2} \, \text{m} \): \[ F_1 = \frac{G \cdot 1 \cdot 2}{(\sqrt{2})^2} = \frac{G \cdot 2}{2} = G \] Thus, the force exerted by each 1 kg mass on the 2 kg mass is \( G \). 3. **Determine the Direction of the Forces**: Since the triangle is equilateral, the angles between the forces exerted by the three masses at the centroid will be 120 degrees. Therefore, we have three forces \( F_1, F_2, F_3 \) acting at angles of 120 degrees to each other. 4. **Use Lami's Theorem**: According to Lami's theorem, if three forces are acting at a point and are in equilibrium, the following relationship holds: \[ \frac{F_1}{\sin(\alpha)} = \frac{F_2}{\sin(\beta)} = \frac{F_3}{\sin(\gamma)} \] Here, \( \alpha = \beta = \gamma = 120^\circ \). Since all forces are equal (\( F_1 = F_2 = F_3 = G \)), we can conclude that: \[ \frac{G}{\sin(120^\circ)} = \frac{G}{\sin(120^\circ)} = \frac{G}{\sin(120^\circ)} \] This confirms that the net force acting on the 2 kg mass is zero. 5. **Conclusion**: Since the net force acting on the mass of 2 kg at the centroid O is zero due to the symmetry and equal magnitudes of the forces, we conclude: \[ \text{Net Force} = 0 \, \text{N} \] ### Final Answer: The force acting on the mass of 2 kg is **0 N**.
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